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#3 2012-11-29 20:15:16
- bobbym
- Administrator
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Re: Indefinite integration
Hi jacks;
The answer for that is huge, are you supposed to do that by hand?
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#4 2012-11-30 00:46:46
- anonimnystefy
- Real Member
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Re: Indefinite integration
bob bundy wrote:
(x + 1) is a factor of top and bottom.
Bob
It is not.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#5 2012-11-30 02:19:20
- Agnishom
- Super Member
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Re: Indefinite integration
It is a factor of the denominator only.
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#7 2012-11-30 05:29:25
- anonimnystefy
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Re: Indefinite integration
The integral can be solved through partial fractions, but it is a nasty factorisation!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#8 2012-11-30 08:20:54
- bobbym
- Administrator
Online
Re: Indefinite integration
Hi;
Yes, it looks like a handmade one. That can be dangerous.
In mathematics, you don't understand things. You just get used to them.
Probability is the most important concept in modern science, especially as nobody has the slightest notion what it means.
90% of mathematicians do not understand 90% of currently published mathematics.
#9 2012-11-30 14:38:17
- noelevans
- Full Member
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Re: Indefinite integration
Hi! 
The third degree polynomial divided by the fifth degree polynomial can be written as an infinite
series as:
Then just integrate it term by term to get the indefinite integral in an infinite series form.
Starting with the 1/x^2 term the coefficients in the series repeat in blocks of 10 in the pattern
1,-1,1,0,0,-1,1,-1,0,0. That is what the 1/(x^10n) handles. So the next 6 terms in the series
would have exponents on the bottom of 12,13,14,17,18,19 and the next 6 would have
22,23,24,27,28,29 for the exponents on the bottom.
The signs in each block of six terms would continue to be 1, -1, 1, -1, 1, -1.
Has anyone got a closed form for the integral? I'd love to see it. If x^5+1 is divided by x-1 I get
x(x+1)(x-1)(x^2+1) + (x+1)/(x-1) but I haven't been able to use this to get a closed form.
Good luck with it!
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